a) 

\(n_{H_2}=\dfrac{0,6}{2}=0,3\left(mol\right)\)

PTHH: 2M + 3H₂SO₄ --> M₂(SO₄)₃ + 3H₂

           0,2<----0,3<--------0,1<-------0,3

=> \(M_M=\dfrac{5,4}{0,2}=27\left(g/mol\right)\)

=> M là Al

b) \(C\%_{dd.H_2SO_4}=\dfrac{0,3.98}{395,2}.100\%=7,44\%\)

c) 

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

          0,2-->0,6

=> \(V_{dd.HCl}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)

a) n_A = m/M_A = 5,4/M_A (mol)

n_H2 = m/M = 0,6/2 = 0,3 (mol)

PTHH:

2A + 3H₂SO₄ (l) → A₂(SO₄)₃ + 3H₂

2 mol : 3 mol : 1 mol : 3 mol

0,2 mol : 0,3 mol : 0,1 mol : 0,3 mol

M_A = m/n = 5,4/0,2 = 27 (g/mol)

=> M là kim loại Al

b) m_H2SO4 = n.M = 0,3.98 = 29,4 (g)

C%_{dd H2SO4} = (m_ct/m_dd).100% = (29,4/395,2).100% ≃ 7,434%

a) PTHH : 2M + 3H2SO4 --> M2(SO4)3 +3H2

n_M =5,4/M_M(mol)

n_H2=0,6/2=0,3(mol)

theo pthh ta có : n_M=2/3nH2=0,2(mol)

=> 5,4/M_M=0,2 => M_M =5,4/0,2=27(g/mol)=> M : Al

b) 2Al + 3H2SO4--> Al2(SO4)3 + 3H2

theo PTHH ta có : n_H2SO4=3/2n_Al=0,3(mol)

=>m_H2SO4=0,3.98=29,4(g)

=> C%dd_H2SO4=29,4/395,2 .100=7,44(%)

\(n_{XCl_3}=\dfrac{a}{M_X+106,5}\left(mol\right)\)

PTHH: 2X + 6HCl --> 2XCl₃ + 3H₂

=> \(n_X=\dfrac{a}{M_X+106,5}\left(mol\right)\)

\(n_{X_2\left(SO_4\right)_3}=\dfrac{b}{2.M_X+288}\left(mol\right)\)

PTHH: 2X + 3H₂SO₄ --> X₂(SO₄)₃ + 3H₂

=> \(n_X=\dfrac{b}{M_X+144}\left(mol\right)\)

\(a,n_{H_2SO_4}=0,3.0,75+0,3.0,25=0,3\left(mol\right)\\ V_{ddH_2SO_4}=300+300=600\left(ml\right)=0,6\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,6}=0,5M\\ m_{H_2SO_4}=0,3.98=29,4\left(g\right)\\ m_{ddH_2SO_4}=600.1,02=612\left(g\right)\\ \rightarrow C\%_{H_2SO_4}=\dfrac{29,4}{612}.100\%=4,8\%\)

\(b,\) Đặt kim loại M có hoá trị n (n ∈ N*)

PTHH: \(2M+nH_2SO_4\rightarrow M_2\left(SO_4\right)_n+nH_2\uparrow\)

           \(\dfrac{0,6}{n}\)<---0,3--------------------------->0,3

\(\rightarrow M_M=\dfrac{5,4}{\dfrac{0,6}{n}}=9n\left(g\text{/}mol\right)\)

Vì n là hoá trị của M nên ta xét bảng

Vậy M là Al

\(c,n_{KClO_3}=\dfrac{15,3125}{122,5}=0,125\left(mol\right)\)

PTHH:

\(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

0,3-->0,15

\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\uparrow\)

0,1<---------------------0,15

\(\rightarrow H=\dfrac{0,1}{0,125}.100\%=80\%\)